Lines and Planes in Space: Learn It 5

Parallel and Intersecting Planes

Unlike lines, which can be parallel, intersecting, or skew in 3D space, planes have only two possible relationships: they are either parallel or they intersect. There’s no such thing as skew planes.

Parallel planes have parallel normal vectors. If two planes have normal vectors [latex]\mathbf{n}_1[/latex] and [latex]\mathbf{n}_2[/latex], the planes are parallel when [latex]\mathbf{n}_1 = k\mathbf{n}_2[/latex] for some scalar [latex]k[/latex].

Intersecting planes meet along a line. When two planes aren’t parallel, they must intersect, and their intersection is always a straight line.

This figure is two planes that are intersecting. The intersection forms a line segment.
Figure 5. The intersection of two nonparallel planes is always a line.
Why Always a Line?: Think about it geometrically. Two distinct planes can’t intersect at just one point (that would require them to be identical), and they can’t intersect at scattered points. The intersection must be a continuous set of points that forms a line.

When two planes intersect, you can find parametric equations for their line of intersection by using both plane equations simultaneously.

Find parametric and symmetric equations for the line formed by the intersection of the planes given by [latex]x+y+z=0[/latex] and [latex]2x-y+z=0[/latex] (see the following figure).

This figure is two planes intersecting in the 3-dimensional coordinate system.
Figure 6. The line formed by the intersection of the planes given by [latex]x+y+z=0[/latex] and [latex]2x-y+z=0[/latex]

Finding the Angle Between Planes

When planes intersect, you might need to find the angle between them. We can use normal vectors to calculate the angle between the two planes.

The key insight: the angle between two planes equals the angle between their normal vectors.

This figure is two parallelograms representing planes. The planes intersect forming angle theta between them. The first plane as vector “n sub 1” normal to the plane. The second vector has vector “n sub 2” normal to the plane. The normal vectors intersect and form the angle theta.
Figure 7. The angle between two planes has the same measure as the angle between the normal vectors for the planes.

We can find the measure of the angle [latex]\theta[/latex] between two intersecting planes by first finding the cosine of the angle, using the following equation:

[latex]\large{\cos\theta=\frac{|{\bf{n}}_1\cdot{\bf{n}}_2|}{||{\bf{n}}_1|| \ ||{\bf{n}}_2||}}[/latex].

We can then use the angle to determine whether two planes are parallel or orthogonal or if they intersect at some other angle.

angle between planes

For planes with normal vectors [latex]\mathbf{n}_1[/latex] and [latex]\mathbf{n}_2[/latex], the angle [latex]\theta[/latex] between the planes is:

[latex]\cos \theta = \frac{|\mathbf{n}_1 \cdot \mathbf{n}_2|}{|\mathbf{n}_1| , |\mathbf{n}_2|}[/latex]

We use absolute value to ensure [latex]0° \leq \theta \leq 90°[/latex].

[latex]\\[/latex]

Special Cases:

  • Parallel planes: [latex]\mathbf{n}_1 \times \mathbf{n}_2 = \mathbf{0}[/latex] (cross product is zero)
  • Perpendicular planes: [latex]\mathbf{n}_1 \cdot \mathbf{n}_2 = 0[/latex] (dot product is zero)

Determine whether each pair of planes is parallel, orthogonal, or neither. If the planes are intersecting, but not orthogonal, find the measure of the angle between them. Give the answer in radians and round to two decimal places.

  1. [latex]x+2y-z=8[/latex] and [latex]2x+4y-2z=10[/latex]
  2. [latex]2x-3y+2z=3[/latex] and [latex]6x+2y-3z=1[/latex]
  3. [latex]x+y+z=4[/latex] and [latex]x-3y+5z=1[/latex]

Distance Between Parallel Planes

When two planes are parallel, you can find the distance between them by selecting any point on one plane and calculating its distance to the other plane.

distance from point to plane

The distance from point [latex]P(x_0, y_0, z_0)[/latex] to plane [latex]ax + by + cz + k = 0[/latex] is:

[latex]d = \frac{|ax_0 + by_0 + cz_0 + k|}{\sqrt{a^2 + b^2 + c^2}}[/latex]

How This Works

If [latex](x_1, y_1, z_1)[/latex] is any point on the plane [latex]ax + by + cz + k = 0[/latex], then [latex]ax_1 + by_1 + cz_1 + k = 0[/latex], so [latex]k = -ax_1 - by_1 - cz_1[/latex].

Using our general distance formula:

[latex]d = \frac{|a(x_0-x_1) + b(y_0-y_1) + c(z_0-z_1)|}{\sqrt{a^2+b^2+c^2}}[/latex]

Expanding the numerator:

[latex]d = \frac{|ax_0 + by_0 + cz_0 - ax_1 - by_1 - cz_1|}{\sqrt{a^2+b^2+c^2}} = \frac{|ax_0 + by_0 + cz_0 + k|}{\sqrt{a^2+b^2+c^2}}[/latex]

For parallel planes [latex]ax + by + cz + k_1 = 0[/latex] and [latex]ax + by + cz + k_2 = 0[/latex], the distance between them is [latex]\frac{|k_1 - k_2|}{\sqrt{a^2 + b^2 + c^2}}[/latex]. Just pick the origin as your test point!
Find the distance between the two parallel planes given by [latex]2x+y-z=2[/latex] and [latex]2x+y-z=8[/latex].